Question

1) Tính:

\(\frac{7.8^3-5.2^{10}}{\left(-16\right)^2}\)

2)Tìm x:

a)\(\frac{108}{12}\le x\le\frac{91}{7}\)

b)\(\frac{-28}{4}\le x\le\frac{-21}{7}\)

Answer 1

1)

\(\frac{7.8^3-5.2^{10}}{\left(-16\right)^2}\)       

=  \(\frac{7.2^8.2-5.2^8.2^2}{16^2}\)

=  \(\frac{2^8.\left(2.7-5.2^2\right)}{2^8}\)

=   \(\frac{2^8.\left(-6\right)}{2^8}\)

=   \(-6\)

 

Answer 2

2)

b)\(\frac{-28}{4}\le x\le\frac{-21}{7}\)

\(\Rightarrow-7\le x\le-3\)

\(\Rightarrow x=\left\{-7;-6;-5;-4;-3\right\}\)

Answer 3

2)

a)\(\frac{108}{12}\le x\le\frac{91}{7}\)

\(\Rightarrow9\le x\le13\)

\(\Rightarrow x=\left\{9;10;11;12;13\right\}\)

Answer 4

-6